Difference between revisions of "2005 AIME II Problems/Problem 9"
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We know by [[De Moivre's Theorem]] that <math>(\cos t + i \sin t)^n = \cos nt + i \sin nt</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. | We know by [[De Moivre's Theorem]] that <math>(\cos t + i \sin t)^n = \cos nt + i \sin nt</math> for all [[real number]]s <math>t</math> and all [[integer]]s <math>n</math>. So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem. | ||
− | Recall the [[trigonometric identities]] <math>\cos \frac{\pi}2 - u = \sin u</math> and <math>\sin \frac{\pi}2 - u = \cos u</math> hold for all real <math>u</math>. If our original equation holds for all <math>t</math>, it must certainly hold for <math>t = \frac{\pi}2 - u</math>. Thus, the question is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</math> we have that <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)</math> holds for all real <math>u</math>. | + | Recall the [[trigonometric identities]] <math>\cos \left(\frac{\pi}2 - u\right) = \sin u</math> and <math>\sin \left(\frac{\pi}2 - u\right) = \cos u</math> hold for all real <math>u</math>. If our original equation holds for all <math>t</math>, it must certainly hold for <math>t = \frac{\pi}2 - u</math>. Thus, the question is equivalent to asking for how many [[positive integer]]s <math>n \leq 1000</math> we have that <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \sin n \left(\frac\pi2 -u \right) + i\cos n \left(\frac\pi2 - u\right)</math> holds for all real <math>u</math>. |
<math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu</math>. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all <math>n</math> such that <math>\cos n u = \sin n\left(\frac\pi2 - u\right)</math> and <math>\sin nu = \cos n\left(\frac\pi2 - u\right)</math> hold for all real <math>u</math>. | <math>\left(\sin\left(\frac\pi2 - u\right) + i \cos\left(\frac\pi 2 - u\right)\right)^n = \left(\cos u + i \sin u\right)^n = \cos nu + i\sin nu</math>. We know that two [[complex number]]s are equal if and only if both their [[real part]] and [[imaginary part]] are equal. Thus, we need to find all <math>n</math> such that <math>\cos n u = \sin n\left(\frac\pi2 - u\right)</math> and <math>\sin nu = \cos n\left(\frac\pi2 - u\right)</math> hold for all real <math>u</math>. | ||
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This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</math>. Notice, <math>\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}</math> since <math>\sin(-t) = -\sin(t)</math>. Using this, <math>(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)</math> is recast as <math>(i e^{-it})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> is a multiple of <math>4</math> exactly one quarter of the residues are congruent to <math>1</math> hence we have <math>\boxed{250}</math>. | This problem begs us to use the familiar identity <math>e^{it} = \cos(t) + i \sin(t)</math>. Notice, <math>\sin(t) + i \cos(t) = i(\cos(t) - i \sin(t)) = i e^{-it}</math> since <math>\sin(-t) = -\sin(t)</math>. Using this, <math>(\sin(t) + i \cos(t))^n = \sin(nt) + i \cos(nt)</math> is recast as <math>(i e^{-it})^n = i e^{-itn}</math>. Hence we must have <math>i^n = i \Rightarrow i^{n-1} = 1 \Rightarrow n \equiv 1 \bmod{4}</math>. Thus since <math>1000</math> is a multiple of <math>4</math> exactly one quarter of the residues are congruent to <math>1</math> hence we have <math>\boxed{250}</math>. | ||
− | == | + | === Solution 3 === |
− | {{ | + | We can rewrite <math>\sin(t)</math> as <math>\cos\left(\frac{\pi}{2}-t\right)</math> and <math>\cos(t)</math> as <math>\sin\left(\frac{\pi}{2}-t\right)</math>. This means that <math>\sin t + i\cos t = e^{i\left(\frac{\pi}{2}-t\right)}=\frac{e^{\frac{\pi i}{2}}}{e^{it}}</math>. This theorem also tells us that <math>e^{\frac{\pi i}{2}}=i</math>, so <math>\sin t + i\cos t = \frac{i}{e^{it}}</math>. By the same line of reasoning, we have <math>\sin nt + i\cos nt = \frac{i}{e^{int}}</math>. |
+ | |||
+ | For the statement in the question to be true, we must have <math>\left(\frac{i}{e^{it}}\right)^n=\frac{i}{e^{int}}</math>. The left hand side simplifies to <math>\frac{i^n}{e^{int}}</math>. We cancel the denominators and find that the only thing that needs to be true is that <math>i^n=i</math>. This is true if <math>n\equiv1\pmod{4}</math>, and there are <math>\boxed{250}</math> such numbers between <math>1</math> and <math>1000</math>. Solution by Zeroman | ||
+ | |||
+ | === Solution 4 === | ||
+ | We are using degrees in this solution instead of radians. I just process stuff better that way. | ||
+ | |||
+ | We can see that the LHS is <math>cis(n(90^{\circ}-t))</math>, and the RHS is <math>cis(90^{\circ}-nt)</math> So, <math>n(90-t) \equiv 90-nt \mod 360</math> Expanding and canceling the nt terms, we will get <math>90n \equiv 90 \mod 360</math>. Canceling gets <math>n \equiv 1 \mod 4</math>, and thus there are <math>\boxed{250}</math> values of n. | ||
+ | |||
+ | -AlexLikeMath | ||
+ | |||
+ | === Solution 5(CHEAP) === | ||
+ | Let <math>t=0</math>. Then, we have <math>i^n=i</math> which means <math>n\equiv 1\pmod{4}</math>. Thus, the answer is <math>\boxed{250}</math>. | ||
+ | |||
+ | === Solution 6 === | ||
+ | |||
+ | We factor out <math>i^n</math> from <math>(\sin t + i \cos t)^n = i^n (\cos(t) - i \sin t)= i^n(\cos(nt) - i\sin nt).</math> We know the final expression must be the same as <math>\sin nt + i \cos nt</math> so we must have <math>i^n(\cos(nt) - i\sin nt) = \sin nt + i \cos nt</math> in which testing yields <math>n \equiv 1 \pmod{4}</math> is the only mod that works, so we have a total of <math>1000 \cdot\frac{1}{4} = \boxed{250}</math> integers <math>n</math> that work. | ||
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[[Category:Intermediate Algebra Problems]] | [[Category:Intermediate Algebra Problems]] | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | ==See Also== | ||
+ | {{AIME box|year=2005|n=II|num-b=8|num-a=10}} |
Latest revision as of 14:03, 1 September 2020
Contents
Problem
For how many positive integers less than or equal to is true for all real ?
Solution
Solution 1
We know by De Moivre's Theorem that for all real numbers and all integers . So, we'd like to somehow convert our given expression into a form from which we can apply De Moivre's Theorem.
Recall the trigonometric identities and hold for all real . If our original equation holds for all , it must certainly hold for . Thus, the question is equivalent to asking for how many positive integers we have that holds for all real .
. We know that two complex numbers are equal if and only if both their real part and imaginary part are equal. Thus, we need to find all such that and hold for all real .
if and only if either or for some integer . So from the equality of the real parts we need either , in which case , or we need , in which case will depend on and so the equation will not hold for all real values of . Checking in the equation for the imaginary parts, we see that it works there as well, so exactly those values of congruent to work. There are of them in the given range.
Solution 2
This problem begs us to use the familiar identity . Notice, since . Using this, is recast as . Hence we must have . Thus since is a multiple of exactly one quarter of the residues are congruent to hence we have .
Solution 3
We can rewrite as and as . This means that . This theorem also tells us that , so . By the same line of reasoning, we have .
For the statement in the question to be true, we must have . The left hand side simplifies to . We cancel the denominators and find that the only thing that needs to be true is that . This is true if , and there are such numbers between and . Solution by Zeroman
Solution 4
We are using degrees in this solution instead of radians. I just process stuff better that way.
We can see that the LHS is , and the RHS is So, Expanding and canceling the nt terms, we will get . Canceling gets , and thus there are values of n.
-AlexLikeMath
Solution 5(CHEAP)
Let . Then, we have which means . Thus, the answer is .
Solution 6
We factor out from We know the final expression must be the same as so we must have in which testing yields is the only mod that works, so we have a total of integers that work.
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
See Also
2005 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |